Question

Of the following equilibria, only ________ will shift to the right in response to a decrease in volume. N2 (g) + 3H2 (g) 2NH3 (g) 2 Fe2O3 (s) 4 Fe (s) + 3O2 (g) 2 SO3 (g) 2 SO2 (g) + O2 (g) H2 (g) + Cl2 (g) 2 HCl (g) 2HI (g) H2 (g) + I2 (g)

Answer 1

Answer:

Of the following equilibria, only one will shift to the right in response to a decrease in volume.

$N_2 (g) + 3H_2 (g)\rightleftharpoons 2NH_3 (g)$

On decreasing the volume the equilibrium will shift in right direction due to less number of gaseous moles on product side.

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

Decrease the volume

If the volume of the container is decreased , the pressure will increase according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease in pressure is taking place. So, the equilibrium will shift in the direction number of gaseous moles are less.

$N_2 (g) + 3H_2 (g)\rightleftharpoons 2NH_3 (g)$

On decreasing the volume the equilibrium will shift in right direction due to less number of gaseous moles on product side.

$2 Fe_2O_3 (s) \rightleftharpoons 4 Fe (s) + 3O_2 (g)$

On decreasing the volume the equilibrium will shift in left direction due to less number of gaseous moles on reactant side.

$2 SO_3 (g) \rightleftharpoons 2 SO_2 (g) + O_2 (g)$

On decreasing the volume the equilibrium will shift in left direction due to less number of gaseous moles on reactant side.

$H_2 (g) + Cl_2 (g) \rightleftharpoons 2 HCl (g)$

On decreasing the volume the equilibrium will shift in no direction due to same number of gaseous moles on both sides.

$2HI (g) \rightleftharpoons H_2 (g) + I_2 (g)$

On decreasing the volume the equilibrium will shift in no direction due to same number of gaseous moles on both sides.